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            <ol class="post-toc"><li class="post-toc-item post-toc-level-2"><a class="post-toc-link" href="#什么是最大似然估计？"><span class="post-toc-text">什么是最大似然估计？</span></a></li><li class="post-toc-item post-toc-level-2"><a class="post-toc-link" href="#为什么要有参数估计？"><span class="post-toc-text">为什么要有参数估计？</span></a></li><li class="post-toc-item post-toc-level-2"><a class="post-toc-link" href="#举例"><span class="post-toc-text">举例</span></a></li><li class="post-toc-item post-toc-level-2"><a class="post-toc-link" href="#总结"><span class="post-toc-text">总结</span></a></li></ol>
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        <h1 class="post-card-title">最大似然估计(MLE)</h1>
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            <time class="post-time" title="2019-06-02 18:23:12" datetime="2019-06-02T10:23:12.000Z" itemprop="datePublished">


2019-06-02

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	<ul class="article-category-list">作者：吴光生</ul>



            
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            <h2 id="什么是最大似然估计？"><a href="#什么是最大似然估计？" class="headerlink" title="什么是最大似然估计？"></a>什么是最大似然估计？</h2><p>最大似然估计（Maximum Likelihood Estimation）是一种参数估计的方法。其核心思想是：找到参数$\theta$的一个估计值，使得当前样本出现的可能性最大。</p>
 <a id="more"></a>
<p>设总体$X$属于离散型，其分布律$P\{ X=x\}=p(x;\theta), \theta \in \Theta$的形式为已知，$\theta$为待估参数，$\Theta$是$\theta$可能取值的范围。设$X_1,X_2,\cdots,X_n$是来自$X$的样本，则$X_1,X_2,\cdots,X_n$的联合分布律为：</p>
<script type="math/tex; mode=display">
\prod_{i=1}^{n}p(x_i;\theta)</script><p>样本$X_1,X_2,\cdots,X_n$取到观察值$x_1,x_2,\cdots,x_n$的概率，即事件$\{X_1=x_1,X_2=x_2,\cdots,X_n=x_n\}$发生的概率为：</p>
<script type="math/tex; mode=display">
L(\theta)=L(x_1,x_2,\cdots,x_n;\theta)=\prod_{i=1}^{n}p(x_i;\theta),\theta\in\Theta</script><p>这一概率随$\theta$的取值而变化，它是$\theta$的函数，$L(\theta)$称为样本的<strong>似然函数</strong>。</p>
<p>最大似然估计法，就是固定样本观察值$x_1,x_2,\cdots,x_n$，在参数$\theta$取值的可能范围$\Theta$内挑选出一个$\hat\theta$，使得似然函数$L(x_1,x_2,\cdots,x_n;\theta)$达到最大。即：</p>
<script type="math/tex; mode=display">
\hat\theta=\mathop{\arg\max}_{\theta\in\Theta}L(x_1,x_2,\cdots,x_n;\theta)</script><p>这样得到的$\hat\theta$与样本值$x_1,x_2,\cdots,x_n$有关，常记为$\hat\theta(x_1,x_2,\cdots,x_n)$，称为参数$\theta$的<strong>最大似然估计值</strong>，而相应的统计量$\hat\theta(X_1,X_2,\cdots,X_n)$称为参数$\theta$的<strong>最大似然估计量</strong>。</p>
<h2 id="为什么要有参数估计？"><a href="#为什么要有参数估计？" class="headerlink" title="为什么要有参数估计？"></a>为什么要有参数估计？</h2><p>当模型已定，参数未知时。</p>
<p>例如，假设我们知道全国人民的身高服从正态分布，但不知道均值和方差。这时可以通过采样，观察其结果，然后再用样本数据的结果推出正态分布的均值与方差的最大概率值，这样就可以知道全国人民的身高分布的函数。</p>
<h2 id="举例"><a href="#举例" class="headerlink" title="举例"></a>举例</h2><p><span>1. </span> 抛硬币。现有一个正反面不是很均匀的硬币，如果正面朝上记为H，反面朝上记为T，抛10次的结果如下：</p>
<p>   T, H, T, T, H, T, T, T, H, T</p>
<p>   求这个硬币正面朝上的概率有多大？</p>
<p>   很显然，这个概率是0.3。现在用MLE的思想来求解它。</p>
<p>   设$x_1,x_2,\cdots,x_n$是相应于样本$X_1,X_2,\cdots,X_n$的一个样本值。</p>
<p>   不妨用$x_i=1$表示正面朝上，$x_i=0$表示反面朝上</p>
<p>   设正面朝上的概率为$\theta$，抛硬币服从二项分布$X \sim b(1,\theta)$，$X$的分布律为：</p>
<script type="math/tex; mode=display">
   P\{X=x\}=p(x;\theta)=\theta^x (1-\theta)^{1-x},    x=0,1</script><p>   似然函数为：</p>
<script type="math/tex; mode=display">
   L(\theta)=\prod_{i=1}^{n}p(x_i;\theta)=\prod_{i=1}^{n}\theta^{x_i}(1-\theta)^{1-x_i}</script><p>   取对数后，为</p>
<script type="math/tex; mode=display">
\begin{align*}
   ln L(\theta)&=ln\prod_{i=1}^{n}\theta^{x_i}(1-\theta)^{1-x_i}\\
   &=\sum_{i=1}^{n}ln[\theta^{x_i}(1-\theta)^{1-x_i}]\\
   &=\sum_{i=1}^{n}[ln\theta^{x_i}+ln(1-\theta)^{1-x_i}]\\
   &=\sum_{i=1}^{n}[x_iln\theta+(1-x_i)ln(1-\theta)]\\
   &=\sum_{i=1}^{n}x_iln\theta+(n-\sum_{i=1}^{n}x_i)ln(1-\theta)
   \end{align*}</script><p>   求导：</p>
<script type="math/tex; mode=display">
   \frac{\partial ln L(\theta)}{\partial \theta}=\frac{\sum_{i=1}^{n} x_i}{\theta}-\frac{n-\sum_{i=1}^{n} x_i}{1-\theta}</script><p>   令$\frac{\partial ln L(\theta)}{\partial \theta}=0$，可得：</p>
<script type="math/tex; mode=display">
\hat \theta = \frac{\sum_{i=1}^{n} x_i}{n}</script><p>   可知概率$\hat \theta=0.3$</p>
<p><span>2. </span> 设$X \sim N(\mu, \sigma^2)$, $\mu, \sigma^2$为未知参数，$x_1,x_2,\cdots,x_n$是来自$X$的一个样本值。求$\mu, \sigma^2$的最大似然估计量。</p>
<p>   解：$X$的概率密度为：</p>
<script type="math/tex; mode=display">
   f(x;\mu,\sigma^2)=\frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x-\mu)^2]</script><p>   似然函数为：</p>
<script type="math/tex; mode=display">
   \begin{align*}
   L(\mu,\sigma^2)&=\prod_{i=1}^n \frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x_i-\mu)^2]\\
   &=(\frac{1}{\sqrt{2\pi}\sigma})^n exp[-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2]\\
   &=(\frac{1}{2\pi \sigma^2})^{\frac{n}{2}} exp[-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2]\\
   &=(2\pi)^{-\frac{n}{2}}(\sigma^2)^{-\frac{n}{2}}exp[-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2]
   \end{align*}</script><p>   它的对数：</p>
<script type="math/tex; mode=display">
   ln L(\mu,\sigma^2)=-\frac{n}{2}ln(2\pi)-\frac{n}{2}ln(\sigma^2)-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu)^2</script><p>   令</p>
<script type="math/tex; mode=display">
   \left\{
       \begin{array}{l}
           \frac{\partial ln L(\mu,\sigma^2)}{\partial \mu}=\frac{1}{\sigma^2}\Sigma_{i=1}^n(x_i-\mu)=0\\

           \frac{\partial ln L(\mu,\sigma^2)}{\partial \sigma^2}= -\frac{n}{2\sigma^2} + \frac{1}{2 \sigma^4}\sum_{i=1}^n (x_i-\mu)^2=0
       \end{array}
   \right.</script><p>   联合求解，得到参数$\mu$和$\sigma^2$的最大似然估计值分别为：</p>
<script type="math/tex; mode=display">
   \left\{
       \begin{array}{l}
           \hat \mu = \overline x = \frac{1}{n} \sum_{i=1}^n x_i\\

           \hat \sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i-\overline x)^2
       \end{array}
   \right.</script><p>   相应的最大似然估计量分别为：    $\hat \mu=\bar X$,       $\hat \sigma^2=\frac{1}{n}\sum_{i=1}^n (X_i- \overline X)$</p>
<h2 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h2><p>求最大似然估计值的一般步骤：</p>
<p>（1）写出总体$X$的分布律$p(x; \theta)$（$X$为离散型随机变量）或者概率密度$f(x; \theta)$（$X$为连续型随机变量）</p>
<p>（2）写出样本的似然函数$L(\theta)$ </p>
<p>离散型：$L(\theta)=L(x_1,x_2,\cdots,x_n;\theta)=\prod_{i=1}^{n}p(x_i;\theta)$<br>连续型：$L(\theta)=L(x_1,x_2,\cdots,x_n;\theta)=\prod_{i=1}^{n}f(x_i;\theta)$</p>
<p>（3）对似然函数取对数     $ln L(\theta)$</p>
<p>（4）求偏导     $\frac{\partial ln L(\theta)}{\partial \theta}$</p>
<p>（5）解方程（组） $\frac{\partial ln L(\theta)}{\partial \theta}=0$，得到参数$\theta$的最大似然估计值$\hat \theta$</p>
<p>注：参数可能是一个（如例1，只有一个参数$\theta$），也可能是一组（如例2，有两个参数：$\mu, \sigma^2$），一组参数时，求解方法类似。</p>
<p><br></p>
<p><strong>参考文献：</strong></p>
<p>[1] 盛骤, 谢式千, 潘承毅.  概率论与数理统计（第四版）[M].  北京: 高等教育出版社, 2008.</p>

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